What is the solution for y^{''}-5y^'-14y=e^x(x+1) ?

The solution for y^{''}-5y^'-14y=e^x(x+1) is y=c_{1}e^{7x}+c_{2}e^{-2x}-(e^xx)/(18)-5/108 e^x

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y^{''}-5y^'-14y=e^x(x+1)

y^{^{''}} - 5 y^{^{'}} - 14 y = e^{x} (x + 1)

y = c_{1} e^{7 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{18} - \frac{5}{108} e^{x}

Solution steps

y^{''} (x) - 5 y^{'} (x) - 14 y = e^{x} (x + 1)

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{18} - \frac{5}{108} e^{x}

y = c_{1} e^{7 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{18} - \frac{5}{108} e^{x}

y^{^{''}} + 4 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

(D^{3} + D^{2} + D + 1) y = x e^{x}

30 y^{^{''}} + 25 y^{^{'}} + 4 y = 40

y^{^{''}} + y^{^{'}} - 12 y = 4

y^{^{''}} - y = sec^{3} (x) - sec (x)

What is the solution for y^{''}-5y^'-14y=e^x(x+1) ?
The solution for y^{''}-5y^'-14y=e^x(x+1) is y=c_{1}e^{7x}+c_{2}e^{-2x}-(e^xx)/(18)-5/108 e^x